SOLUTION: question:
an executive committee consisted of ten members: four women and six men. Three members were selected at random to be sent to a metting in Hawaii. A blindfolded woman
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Question 322522: question:
an executive committee consisted of ten members: four women and six men. Three members were selected at random to be sent to a metting in Hawaii. A blindfolded woman drew three of the ten names from a hat. All three names drawn were's women's. What was the probability of such luck?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The probability of 
successes in 
trials where 
is the probability of success on any given trial is given by:
\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k})
Where )
is the number of combinations of things taken at a time and is calculated by !})
Here you need 3 successes in 3 trials where the probability of success is 0.4 (4 out of 10 names were women)
\ =\ \left(3\cr 3\right\)(0.4)^3(0.6)^{0})
The arithmetic is yours to do. Hint:
\ =\ 1\ \forall\ n\ \in\ \mathbb{Z},\ n\ >\ 0)
and
John
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